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Question
A solid sphere of mass 0⋅50 kg is kept on a horizontal surface. The coefficient of static friction between the surfaces in contact is 2/7. What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface?
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Solution
Let α be the angular acceleration produced in the sphere.
Rotational equation of motion,
\[F \times R - f_r \times R = I\alpha\]
\[\Rightarrow F = \frac{2}{5}mR\alpha + \mu mg........(1)\]
Translational equation of motion,
\[F = ma - \mu mg\]
\[ \Rightarrow a = \frac{\left( F + \mu mg \right)}{m}\]
For pure rolling, we have
\[\alpha = \frac{a}{R}\]
\[\Rightarrow \alpha = \frac{\left( F + \mu mg \right)}{mR}\]
Putting the value of \[\alpha\] in equation (1), we get
\[F = \frac{2}{5}\frac{mR\left( F + \mu mg \right)}{mR} + \mu mg\]
\[ \Rightarrow F = \frac{2}{5}\left( F + \mu mg \right) \mu mg\]
\[ \Rightarrow F = \frac{2}{5}F + \left( \frac{2}{5} \times \frac{2}{7} \times 0 . 5 \times 10 \right) + \left( \frac{2}{7} \times 0 . 5 \times 10 \right)\]
\[ \Rightarrow \frac{3F}{5} = \frac{4}{7} + \frac{10}{7} = 2\]
\[ \Rightarrow F = \frac{5 \times 2}{3} = \frac{10}{3} = 3 . 3 N\]
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